New predictions when model contains a poly term
I am trying to predict y for a new vector of data x_new by hand. By "by hand" I mean not using the predict function (my actual model is an mcmc object which predict doesn't accept). This is fine with a simple model like this:
lm1 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + Sepal.Length, data=iris)
x_new <- c(1, 1.4, 3.2, 5.2)
y <- x_new %*% lm1$coef
But I'm not sure how to proceed when my model looks like:
lm2 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + poly(Sepal.Length,3), data=iris)
How exactly do I use the parameters from the poly() variable?
r regression linear-regression prediction predict
asked Nov 24 '18 at 22:20
John FJohn F
19211
add a comment |
I am trying to predict y for a new vector of data x_new by hand. By "by hand" I mean not using the predict function (my actual model is an mcmc object which predict doesn't accept). This is fine with a simple model like this:
lm1 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + Sepal.Length, data=iris)
x_new <- c(1, 1.4, 3.2, 5.2)
y <- x_new %*% lm1$coef
But I'm not sure how to proceed when my model looks like:
lm2 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + poly(Sepal.Length,3), data=iris)
How exactly do I use the parameters from the poly() variable?
r regression linear-regression prediction predict
asked Nov 24 '18 at 22:20
John FJohn F
19211
add a comment |
I am trying to predict y for a new vector of data x_new by hand. By "by hand" I mean not using the predict function (my actual model is an mcmc object which predict doesn't accept). This is fine with a simple model like this:
lm1 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + Sepal.Length, data=iris)
x_new <- c(1, 1.4, 3.2, 5.2)
y <- x_new %*% lm1$coef
But I'm not sure how to proceed when my model looks like:
lm2 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + poly(Sepal.Length,3), data=iris)
How exactly do I use the parameters from the poly() variable?
r regression linear-regression prediction predict
asked Nov 24 '18 at 22:20
John FJohn F
19211
I am trying to predict y for a new vector of data x_new by hand. By "by hand" I mean not using the predict function (my actual model is an mcmc object which predict doesn't accept). This is fine with a simple model like this:
lm1 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + Sepal.Length, data=iris)
x_new <- c(1, 1.4, 3.2, 5.2)
y <- x_new %*% lm1$coef
But I'm not sure how to proceed when my model looks like:
lm2 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + poly(Sepal.Length,3), data=iris)
How exactly do I use the parameters from the poly() variable?
r regression linear-regression prediction predict
r regression linear-regression prediction predict
asked Nov 24 '18 at 22:20
John FJohn F
19211
asked Nov 24 '18 at 22:20
John FJohn F
19211
asked Nov 24 '18 at 22:20
John FJohn F
19211
asked Nov 24 '18 at 22:20
John FJohn F
19211
asked Nov 24 '18 at 22:20
John FJohn F
19211
19211
add a comment |
add a comment |
1 Answer
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You need to set poly(., raw=TRUE) so it uses raw rather than orthogonal polynomials. Compare, now they yield the same coefficients:
lm2 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + Sepal.Length +
I(Sepal.Length^2) + I(Sepal.Length^3), data=iris)
lm3 <- lm(Petal.Width ~ Petal.Length + poly(Sepal.Length, 3, raw=TRUE),
data=iris)
> coef(lm2)
(Intercept) Petal.Length Sepal.Width Sepal.Length I(Sepal.Length^2) I(Sepal.Length^3)
10.22126962 0.50889848 0.22999328 -5.81536464 0.98349473 -0.05626378
> coef(lm3)
(Intercept) Petal.Length Sepal.Width poly(Sepal.Length, 3, raw = TRUE)1
10.22126962 0.50889848 0.22999328 -5.81536464
poly(Sepal.Length, 3, raw = TRUE)2 poly(Sepal.Length, 3, raw = TRUE)3
0.98349473 -0.05626378
answered Nov 24 '18 at 22:33
jay.sfjay.sf
4,61321439
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
You need to set poly(., raw=TRUE) so it uses raw rather than orthogonal polynomials. Compare, now they yield the same coefficients:
lm2 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + Sepal.Length +
I(Sepal.Length^2) + I(Sepal.Length^3), data=iris)
lm3 <- lm(Petal.Width ~ Petal.Length + poly(Sepal.Length, 3, raw=TRUE),
data=iris)
> coef(lm2)
(Intercept) Petal.Length Sepal.Width Sepal.Length I(Sepal.Length^2) I(Sepal.Length^3)
10.22126962 0.50889848 0.22999328 -5.81536464 0.98349473 -0.05626378
> coef(lm3)
(Intercept) Petal.Length Sepal.Width poly(Sepal.Length, 3, raw = TRUE)1
10.22126962 0.50889848 0.22999328 -5.81536464
poly(Sepal.Length, 3, raw = TRUE)2 poly(Sepal.Length, 3, raw = TRUE)3
0.98349473 -0.05626378
answered Nov 24 '18 at 22:33
jay.sfjay.sf
4,61321439
add a comment |
You need to set poly(., raw=TRUE) so it uses raw rather than orthogonal polynomials. Compare, now they yield the same coefficients:
lm2 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + Sepal.Length +
I(Sepal.Length^2) + I(Sepal.Length^3), data=iris)
lm3 <- lm(Petal.Width ~ Petal.Length + poly(Sepal.Length, 3, raw=TRUE),
data=iris)
> coef(lm2)
(Intercept) Petal.Length Sepal.Width Sepal.Length I(Sepal.Length^2) I(Sepal.Length^3)
10.22126962 0.50889848 0.22999328 -5.81536464 0.98349473 -0.05626378
> coef(lm3)
(Intercept) Petal.Length Sepal.Width poly(Sepal.Length, 3, raw = TRUE)1
10.22126962 0.50889848 0.22999328 -5.81536464
poly(Sepal.Length, 3, raw = TRUE)2 poly(Sepal.Length, 3, raw = TRUE)3
0.98349473 -0.05626378
answered Nov 24 '18 at 22:33
jay.sfjay.sf
4,61321439
add a comment |
You need to set poly(., raw=TRUE) so it uses raw rather than orthogonal polynomials. Compare, now they yield the same coefficients:
lm2 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + Sepal.Length +
I(Sepal.Length^2) + I(Sepal.Length^3), data=iris)
lm3 <- lm(Petal.Width ~ Petal.Length + poly(Sepal.Length, 3, raw=TRUE),
data=iris)
> coef(lm2)
(Intercept) Petal.Length Sepal.Width Sepal.Length I(Sepal.Length^2) I(Sepal.Length^3)
10.22126962 0.50889848 0.22999328 -5.81536464 0.98349473 -0.05626378
> coef(lm3)
(Intercept) Petal.Length Sepal.Width poly(Sepal.Length, 3, raw = TRUE)1
10.22126962 0.50889848 0.22999328 -5.81536464
poly(Sepal.Length, 3, raw = TRUE)2 poly(Sepal.Length, 3, raw = TRUE)3
0.98349473 -0.05626378
answered Nov 24 '18 at 22:33
jay.sfjay.sf
4,61321439
You need to set poly(., raw=TRUE) so it uses raw rather than orthogonal polynomials. Compare, now they yield the same coefficients:
lm2 <- lm(Petal.Width ~ Petal.Length + Sepal.Width + Sepal.Length +
I(Sepal.Length^2) + I(Sepal.Length^3), data=iris)
lm3 <- lm(Petal.Width ~ Petal.Length + poly(Sepal.Length, 3, raw=TRUE),
data=iris)
> coef(lm2)
(Intercept) Petal.Length Sepal.Width Sepal.Length I(Sepal.Length^2) I(Sepal.Length^3)
10.22126962 0.50889848 0.22999328 -5.81536464 0.98349473 -0.05626378
> coef(lm3)
(Intercept) Petal.Length Sepal.Width poly(Sepal.Length, 3, raw = TRUE)1
10.22126962 0.50889848 0.22999328 -5.81536464
poly(Sepal.Length, 3, raw = TRUE)2 poly(Sepal.Length, 3, raw = TRUE)3
0.98349473 -0.05626378
answered Nov 24 '18 at 22:33
jay.sfjay.sf
4,61321439
answered Nov 24 '18 at 22:33
jay.sfjay.sf
4,61321439
answered Nov 24 '18 at 22:33
jay.sfjay.sf
4,61321439
answered Nov 24 '18 at 22:33
jay.sfjay.sf
4,61321439
4,61321439
add a comment |
add a comment |
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